Arrays decay to pointers only when the passing format requests them to decay, i.e. if the corresponding parameter is declared as a pointer (as in this case). If the parameter were declared as a reference to an array, the decay wouldn't take place.
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AnTJul 28 '10 at 17:12

@AndreyT: who said anything about references ? If your function is void func(int a[]) then a will still have decayed to a pointer.
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Paul RJul 28 '10 at 19:13

sizeof returns the size of the type. In the second example, func( int *a ), a is a pointer and sizeof will report it as such. Even if you did func( int a[100] ), a would be a pointer. If you want the size of the array in func, you must pass it as an extra argument.

A pointer is a pointer. That means, it simply points to memory, and that's all about it. Creating a pointer to an array (which usually means a pointer to the first element of the array, but not necessarily) is still only a pointer to some memory location. As a memory address is simply a memory address there is also no way for the pointer to know that the memory it is pointing to originally was an array, or how long that array was.

The function does not know the array size in your example because you took explicit steps to convert your array to pointer type, thus completely stripping the function parameter of its original array nature. Once again, you yourself took deliberate steps to make sure that the function does not know the size of the array. Under these circumstances, it is rather strange to see you ask the question about why the function doesn't know the array size.

If you what the function to receive its argument as an array, you have to pass it as an array, not as a mere pointer, and declare the corresponding function parameter accordingly. Arrays in C++ cannot be passed "by value", which means that you'll have to pass it "by reference", as one possibility